Problem: Divide the following complex numbers. $ \dfrac{-3+15i}{3-2i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${3+2i}$ $ \dfrac{-3+15i}{3-2i} = \dfrac{-3+15i}{3-2i} \cdot \dfrac{{3+2i}}{{3+2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-3+15i) \cdot (3+2i)} {(3-2i) \cdot (3+2i)} = \dfrac{(-3+15i) \cdot (3+2i)} {3^2 - (-2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-3+15i) \cdot (3+2i)} {(3)^2 - (-2i)^2} = $ $ \dfrac{(-3+15i) \cdot (3+2i)} {9 + 4} = $ $ \dfrac{(-3+15i) \cdot (3+2i)} {13} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-3+15i}) \cdot ({3+2i})} {13} = $ $ \dfrac{{-3} \cdot {3} + {15} \cdot {3 i} + {-3} \cdot {2 i} + {15} \cdot {2 i^2}} {13} $ Evaluate each product of two numbers. $ \dfrac{-9 + 45i - 6i + 30 i^2} {13} $ Finally, simplify the fraction. $ \dfrac{-9 + 45i - 6i - 30} {13} = \dfrac{-39 + 39i} {13} = -3+3i $